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    I can do (i) but not (ii). The answer is h^-1(x)=h(x) but I just don't understand it! Thanks!
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    hh(x)=x

    Multiply on the left by h^{-1}

    h^{-1}hh(x)=h^{-1}(x)

    So

    h(x)=h^{-1}(x) since h^{-1}h=1 (identity)
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    (Original post by rpnom)

    I can do (i) but not (ii). The answer is h^-1(x)=h(x) but I just don't understand it! Thanks!
    Just some alternative wording to the above:

    We have h(h(x)) = x. Now you can apply a function to both sides of an equality (if it is injective, for the rigorous) so, for example when you have 2 = 2 you can apply the square function to both sides of the equality to get 2^2 = 2^2.

    Now you have h(h(x)) = x. If we apply the function h^{-1}(x) (that is, the inverse function h) then we get h^{-1}(h(h(x))) = h^{-1}(x).

    But when we apply the inverse function to the function itself, they just cancel out. So for example, when we do the square root of a square, it cancels out. (\sqrt{x})^2 = x. They just cancel each other out, you see? Because squaring and square rooting are inverses of one another.

    So when we apply h^{-1}h(\text{anything}) we get \text{anything} back out.

    In this case, our anything is h(x). So we have h^{-1}h(h(x)) = h(x) since the h^-1 cancels the first h.

    So this means that h^{-1}(h(h(x))) = h^{-1}(x) is really just h(x) = h^{-1}(x).

    Let's recap:

    We had h(h(x)) = x.

    We apply the inverse function h^{-1} to both sides.

    We get h^{-1}(h(h(x))) = h^{-1}(x).

    We simplify the LHS by noting that the inverse function of a function cancels out.

    We get h(x) = h^{-1}(x).
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    (Original post by Zacken)
    Just some alternative wording to the above:

    We have h(h(x)) = x. Now you can apply a function to both sides of an equality (if it is injective, for the rigorous) so, for example when you have 2 = 2 you can apply the square function to both sides of the equality to get 2^2 = 2^2.

    Now you have h(h(x)) = x. If we apply the function h^{-1}(x) (that is, the inverse function h) then we get h^{-1}(h(h(x))) = h^{-1}(x).

    But when we apply the inverse function to the function itself, they just cancel out. So for example, when we do the square root of a square, it cancels out. (\sqrt{x})^2 = x. They just cancel each other out, you see? Because squaring and square rooting are inverses of one another.

    So when we apply h^{-1}h(\text{anything}) we get \text{anything} back out.

    In this case, our anything is h(x). So we have h^{-1}h(h(x)) = h(x) since the h^-1 cancels the first h.

    So this means that h^{-1}(h(h(x))) = h^{-1}(x) is really just h(x) = h^{-1}(x).

    Let's recap:

    We had h(h(x)) = x.

    We apply the inverse function h^{-1} to both sides.

    We get h^{-1}(h(h(x))) = h^{-1}(x).

    We simplify the LHS by noting that the inverse function of a function cancels out.

    We get h(x) = h^{-1}(x).
    Thank you!!!!
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    (Original post by rpnom)
    Thank you!!!!
    No worries.
 
 
 
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